Home / ubuntu / Questions / 161802
Accepted
soandos
Asked: 2012-07-10 17:00:52 +0800 CST

How do I select a field/column from the output of `ls -l`?

  • 772

My goal is deceptively simple (at least to me). I wish to take the output of ls -l or ls -lh and select just one field.

I am looking for this to be as bulletproof as possible, by which I mean, assume that filenames can have a variable number of spaces, not everything in the field has the same length, etc.

Bonus points for having a script that will take the name of the the field (or even just a field number), and then return the contents of the field.

I want to turn

enter image description here

into:

enter image description here

command-line

3 Answers

  1. Best Answer

    Try ls -l | awk '{print $7}'.

    awk selects columns so it's perfect for this task.

    • 59

  2. Never parse ls. Use GNU find. Or if portability isn't important, stat(1).

    find . -maxdepth 1 -printf '%Td\n'

    For reading data other than lists of filenames line-by-line and splitting into fields, see: BashFAQ/001

    There are no methods to reliably read a newline-delimited list of filenames that make sense under most circumstances.

    • 24

  3. You may fetch the specific column in shell like:

    ls -al | while read perm bsize user group size month day time file; do echo $day; done

    or awk as shown in @Corey answer, cut -c44-45 would also work after adjustment (since ls has fixed columns) , or whatever else, however the main problem is that it won't be reliable and bulletproof (e.g. on Unix it may be $6, not $7, and it changes depending on arguments) making it not machine-friendly therefore it is not recommended to parse ls command at all.

    The best is to use different available commands such as find or stat, which can provide relevant options to format the output as you need. For example:

    $ stat -c "%x %n" *
2016-04-10 04:53:07.000000000 +0100 001.txt
2016-04-10 05:08:42.000000000 +0100 7c1c.txt

    To return column of only days of modifications, try this example:

    stat -c "%x" * | while read ymd; do date --date="$ymd" "+%d"; done

    It's worth to note that GNU stat could have different options to BSD stat, so it still won't be bulletproof across different operating systems.

    • 1